Three identical small spheres, each of weight 2 lb, can slide freely on a horizo

Question

Three identical small spheres, each of weight 2 lb, can slide freely on a horizontal frictionless surface. Spheres B and C are connected by a light rod and are at rest in the position shown when sphere B is struck squarely by sphere A which is moving to the right with a velocity v_0 = (8 ft/s)i. Knowing that theta = 45 degree and that the velocities of spheres A and B immediately after the impact are v'_A = 0 and v'_B = (6 ft/s)i + v_B'y j, determine (a) the component v_B'y and (b) the velocity of C immediately after impact.

Explanation / Answer

Let m be the mass of one ball.

Conservation of linear momentum:

(mv)=(mv)0

mVa+mVb+mVc=m(Va)o+m(Vb)o+m(Vc)o

Dividing by m and applying numerical data,

0+[(6ft/s)i+(Vb)yj]+[(Vc)xi+(Vc)yj]=(8ft/s)+0+0

Components:

x:

6+(Vc)x=8

Therefore (Vc)x=2ft/s

y:

(Vb)y+(Vc)y=0

Conservation of angular momentum about O:

[r×(mv)]=[r×(mvo )]

where rA = 0, rB = 0, rC=(1.5 ft)(cos 45i+sin 45j)

(1.5 ft)(cos 45i+sin 45j)×[m(Vc)xi+m(Vc)yj]=0

Since their cross product is zero, the two vectors areparallel.

(Vc)y=(Vc)xtan 45=2 tan45=2ft/s

From (1) (Vb)y=-2ft/s

Vc=(2ft/s)i+(2ft/s)

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